Question

ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that:

(i) ABCD is a square

(ii) Diagonal BD bisects ∠B as well as ∠D.​

Answer 1

Answer:

To Prove :  (i) ABCD is a square.

(ii) Diagonal BD bisects ∠B as  well as ∠D.

Proof : (i) In ∆ABC and ∆ADC, we have  

∠BAC = ∠DAC

[Given]  

∠BCA = ∠DCA

[Given]  AC = AC

∴ ∆ABC ≅ ∆ADC [ASA congruence]  

∴ AB = AD and CB = CD [CPCT]

But, in a rectangle opposite sides are equal,  

i.e., AB = DC and BC = AD  

∴ AB = BC = CD = DA  Hence, ABCD is a square Proved.

(ii) In ∆ABD and ∆CDB, we have

AD = CD  AB = CD [Sides of a square]

BD = BD [Common]  

∴ ∆ABD ≅ ∆CBD [SSS congruence]  

So, ∠ABD = ∠CBD  

∠ADB = ∠CDB  Hence, diagonal BD bisects ∠B as well as ∠D Proved

PLZ MARK AS BRIANLIEST AND THX FOR THE SUPERB QUESTION

Answer 2

SOLUTION:−

Given:-

ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C.

Explanation:-

(i) ∠DAC = ∠DCA (AC bisects ∠A as well as ∠C)

⇒ AD = CD (Sides opposite to equal angles of a triangle are equal)

also, CD = AB (Opposite sides of a rectangle)

AB = BC = CD = AD

Thus, ABCD is a square.

(ii) In ΔBCD,BC = CD

⇒ ∠CDB = ∠CBD (Angles opposite to equal sides are equal)

also, ∠CDB = ∠ABD (Alternate interior angles)

⇒ ∠CBD = ∠ABD

Thus, BD bisects ∠B

Now,

∠CBD = ∠ADB

⇒ ∠CDB = ∠ADB

Thus, BD bisects ∠D